Inside a hydrogen atom

Considering Gauss' law for a spherical charge distribution, it follows $\oint \vec E \cdot d\vec A = E(r) \cdot 4 \pi r^2 = \frac{Q_r}{\varepsilon_0}$ with the charge $Q_r$ , that is enclosed by a sphere with radius $r$ . This charge can be estimated by volume integration with the help of spherical coordinates $\begin{aligned} Q_{r} &= e + \int \rho_{e-}(r') dV = e + \int_0^{r} \int_{0}^{\pi} \int_{0}^{2\pi} \rho_{e-}(r') r'^2 \sin \vartheta d\varphi d\varphi dr' \\ &= e - 4 \pi \int_0^{r} \frac{e}{\pi a^3} e^{-\frac{2 r'}{a}} r'^2 dr = e - \frac{4 e}{a^3} \int_0^{r} \left[\frac{\partial^2}{\partial \alpha^2} e^{-\alpha r'} \right]_{\alpha = \frac{2}{a}} dr' \\ &= e - \frac{4 e}{a^3} \left[ \frac{\partial^2}{\partial \alpha^2} \int_0^{r} e^{-\alpha r'} dr' \right]_{\alpha = \frac{2}{a}} = e - \frac{4 e}{a^3} \left[ \frac{\partial^2}{\partial \alpha^2} \frac{1}{\alpha} \left(1 - e^{-\alpha r} \right) \right]_{\alpha = \frac{2}{a}} \\ &= e - \frac{4 e}{a^3} \left[ \frac{2}{\alpha^3} - \left(\frac{2}{\alpha^3} + \frac{2 r}{\alpha^2} + \frac{r^2}{\alpha} \right) e^{-\alpha r} \right]_{\alpha = \frac{2}{a}} \\ &= e \left(1 + \frac{2 r}{a} + \frac{2 r^2}{a^2} \right) e^{-\frac{2 r}{a}} \end{aligned}$ Therefore, the electrical field for $r = a$ results to $E(a) = \left.\frac{Q_r}{4 \pi \varepsilon_0 r^2}\right|_{r=a} = 5 e^{-2} E_0 \approx 0.677 \cdot E_0$